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It is well known that in characteristic p we have the "freshman dream" $x^p -1 = (x - 1)^p$. Some experimentation seems to suggest that the heuristic computation $(x-1)(x-1)^{p-1} = (x-1)^p = x^p - 1 = (x-1)(1 + x + \dots + x^{p-1}) \implies (x-1)^{p-1} = 1 + x + \dots + x^{p-1}$ actually yields a correct result.

Playing around with binomial coefficients leads me to believe that $ p-1 \choose k$ = $(-1)^k$ mod p, which would imply the result, but I do not see how to show this. It seems like there should be an easy way, I just do not see it.

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You can cancel non-zero elements in integral domains, so there is nothing wrong with just cancelling the $x-1$ factor. –  Ragib Zaman Aug 15 '12 at 14:41
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5 Answers 5

up vote 3 down vote accepted

$\binom{p-1}{0} \equiv 1$ and $\binom{p-1}{k+1} \equiv \binom{p-1}{k} \frac{p-1-k}{k+1} = \binom{p-1}{k} (-1)$ which proves your claim.

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You can multiply and divide (nonzero) numbers mod p, so unless $x \equiv 1 \pmod p$ you can divide both sides of $(x-1)^p \equiv x^p - 1 \equiv (x - 1)(1 + x + ... + x^{p-1}) \pmod p$ by $x - 1$ to get your result. If $x = 1 \pmod p$, then your statement just reduces to $0 \equiv p \pmod p$, which is clear.

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This one is nice! –  Lieven Aug 15 '12 at 14:50
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Hint: show $\frac{x^{p}-1}{x-1}=1+x+...+x^{p-1}$ hence $(x-1)(1+x+...+x^{p-1})=x^{p}-1=(x-1)^{p}$

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You might try:

$$\binom {p-1} k = \binom p {k} -\binom {p-1}{k-1}$$

Since most of the $\binom p k$ are zero taken modulo p the $\binom {p-1}k$ alternate in sign and the values are easy to determine.

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As $p-r≡-r(mod\ p)=>\frac{p-r}{r}≡-1(mod\ p)$

$ p-1 \choose k$ = $\frac{(p-1)(p-2)(p-3)...(p-k)}{1.2.3....k}=\prod_{1 ≤r ≤k} (\frac{p-r}{r})$

Each factor contributing ≡-1(mod p)

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